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प्रश्न
Show that the points (2, 0), (–2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason.
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उत्तर
Let the points be P(2, 0), Q(–2, 0) and R(0, 2).
Distance between two points = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
By distance formula,
d(P, Q) = `sqrt([(-2) - 2]^2 + (0 - 0)^2`
= `sqrt((-4)^2 + (0)^2`
= `sqrt(16 + 0)`
= 4 ...(i)
d(Q, R) = `sqrt([0 - (-2)]^2 + (2 - 0)^2`
= `sqrt((2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(ii)
d(P, R) = `sqrt((0 -2)^2 + (2 - 0)^2`
= `sqrt((- 2)^2 + (2)^2`
= `sqrt(4 + 4)`
= `sqrt(8)` ...(iii)
On adding (ii) and (iii),
d(P, Q) + d(Q, R) = `4 + sqrt(8)`
`4 + sqrt(8) > sqrt(8)`
∴ d(P, Q) + d(Q, R) > d(P, R)
∴ Points P, Q, R are non colinear points.
We can construct a triangle through 3 non collinear points.
∴ The segment joining the given points form a triangle.
Since P(Q, R) = P(P, R)
∴ ∆PQR is an isosceles triangle.
∴ The segment joining the points (2, 0), (–2, 0) and (0, 2) will form an isosceles triangle.
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संबंधित प्रश्न
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
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Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).
Find all possible values of x for which the distance between the points A(x, –1) and B(5, 3) is 5 units.
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Find the value of y for which the distance between the points A (3, −1) and B (11, y) is 10 units.
Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
Find the distance of the following point from the origin :
(5 , 12)
Find the distance of the following point from the origin :
(13 , 0)
Find the relation between x and y if the point M (x,y) is equidistant from R (0,9) and T (14 , 11).
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
Calculate the distance between the points P (2, 2) and Q (5, 4) correct to three significant figures.
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
Find the distance of the following points from origin.
(5, 6)
Find distance between point A(7, 5) and B(2, 5).
Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by ______.
Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.
Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
